Question 15.13: A mass point glides without friction on a cycloid, which is ...
A mass point glides without friction on a cycloid, which is given by x = a(ϑ − sinϑ) and y = a(1+cosϑ) (with 0 ≤ ϑ ≤ 2π). Determine
(a) the Lagrangian, and
(b) the equation of motion.
(c) Solve the equation of motion.
The cycloid is represented by
x = a(ϑ −sinϑ), y = a(1+ cosϑ),where 0 ≤ ϑ ≤ 2π. The kinetic energy is
T = \frac{1}{2} m(\dot{x}^{2} + \dot{y}^{2}) = \frac{1}{2} ma^{2} \left\{ \left[(1 −cos ϑ)\dot{ϑ}\right]^{2} +\left[−(sin ϑ)\dot{ϑ}\right]^{2}\right\},and the potential energy is
V = mgy = mga(1 +cos ϑ).The Lagrangian is given by
L = T −V = ma^{2}(1 −cos ϑ)\dot{ϑ}^{2} −mga(1 + cos ϑ).The equation of motion then reads
\frac{d}{dt} \left(\frac{∂L}{∂ \dot{ϑ}} \right) − \frac{∂L}{∂ϑ} = 0,i.e.,
\frac{d}{dt} \left[ 2ma^{2}(1 − cos ϑ)\dot{ϑ}\right]−\left[ma^{2}(sin ϑ)\dot{ϑ}^{2} + mga sin ϑ\right]= 0or
\frac{d}{dt} [(1 −cos ϑ)\dot{ϑ} ] − \frac{1}{2} (sin ϑ)\dot{ϑ}^{2} − \frac{g}{2a} sin ϑ = 0,i.e.,
(1 −cos ϑ)\ddot{ϑ} + \frac{1}{2} (sinϑ)\dot{ϑ}^{2} − \frac{g}{2a} sin ϑ = 0. (15.35)
By setting u = cos(ϑ/2), one has
\frac{du}{dt} =−\frac{1}{2} sin \left(\frac{ϑ}{2}\right)\dot{ϑ}and
\frac{d^{2}u}{dt^{2}} =−\frac{1}{2} sin \left(\frac{ϑ}{2}\right) \ddot{ϑ} − \frac{1}{4} cos \left(\frac{ϑ}{2}\right) \dot{ϑ}^{2}.Since cot (ϑ/2) = sin ϑ/(1 −cos ϑ), we can write (15.35) as
\ddot{ϑ} + \frac{1}{2} cot \left(\frac{ϑ}{2}\right) \dot{ϑ}^{2} − \frac{g}{2a} cot \left(\frac{ϑ}{2}\right)= 0,and therefore,
\frac{d^{2}u}{dt^{2}} + \frac{g}{4a} u = 0. (15.36)
The solution of this differential equation is
u = cos \left(\frac{ϑ}{2}\right)= C_{1} cos \sqrt{\frac{g}{4a} } t +C_{2} sin \sqrt{\frac{g}{4a}} t.The motion is just like the vibration of an ordinary pendulum of length l = 4a. The arrangement is therefore called a “cycloid pendulum.”